$\begin{aligned} &f(n)=\dfrac{1}{2}n^2-10 \\\\ &h(t)=\dfrac{5-2t}{t-1} \end{aligned}$ $f(h(0.5))=$
Explanation: When evaluating composite functions, we work our way inside out. To evaluate $f(h(0.5))$, let's first evaluate $h(0.5)$. Then we'll plug that result into $f$ to find our answer. Let's evaluate $h({0.5})$. $\begin{aligned}h(t)&=\dfrac{5-2t}{t-1}\\\\ h({0.5})&=\dfrac{5-2({0.5})}{{0.5}-1} ~~~~~~~~~~\text{Plug in }t={0.5}\\\\ &=\dfrac{4}{-0.5}\\\\ &={-8}\end{aligned}$ We now know that $f(h({0.5}))$ is the same as $f({-8})$ because $h({0.5}) = {-8}$. Let's evaluate $f({-8})$. $\begin{aligned}f(n)&=\dfrac{1}{2}n^2-10\\\\ f({{-8}})&=\dfrac{1}{2}({-8})^2-10~~~~~~~~~~\text{Plug in }n={-8}\\\\ &=\dfrac{1}{2}(64)-10\\\\ &=22\\\\\end{aligned}$ The answer: $f(h(0.5))= 22$